Question

A Show that the sum of any three

consecutive even integers is a multiple of 6.
b) Hence, or otherwise, work out what the
values of three consecutive even integers
must be if their sum is 6000.

Answer 1

Below.

Explanation:

a) Let the numbers be n, n + 2 and n + 4

Sum = n + n + 2 + n + 4

= 3n + 6.

As n must be even (2, 4, 6 and so on) , 3n must be divisible by 6 and 3n + 6 is divisible by 6 also.

b) 3n + 6 = 6000

3n = 5994

n = 5994/3 = 1998

So the numbers are 1998, 2000 and 2002.